Question

A graph plotted between $log{t}_{50}$ vs log concentration is a straight line. What conclusion can you draw from the given graph?



[Given: n = order of reaction]

• A. $\text{n}\text{=}\text{1,}{\text{t}}_{\frac{\text{1}}{\text{2}}}\text{=}\frac{\text{1}}{\text{k}\text{.a}}$
• B. $n=2,t_{\frac{1}{2}}=\frac{1}{a}$
• C. $\text{n}\text{=}\text{1,}{\text{t}}_{\frac{\text{1}}{\text{2}}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}$
• D. None of the above

Answer 1

Answer: (C)

$t_{\frac{1}{2}}\propto {\left(a\right)}^{1-n}$

or $t_{\frac{1}{2}}=Z{\left(a\right)}^{{}^{1-n}}$

or $log{t}_{\frac{1}{2}}=logZ+(1-n)loga$

or $(y)=c+mx$

Thus, slope = m = 1 - n

or 1 - n = 0

So n = 1

and for I order reaction $t_{\frac{1}{2}}=\frac{0.693}{k}$ .